I have a sequence $(u_k) \in L^2_{loc}(\mathbb{R}^+; H^1_0(\Omega) )$ and $u \in L^2_{loc}(\mathbb{R}^+\times \Omega )$ such that for any $T >0$ and any compact $K \subset \Omega$ we have : $\int_{[0;T]\times K} |u_k(t,x)-u(t,x)|^2dtdx \rightarrow 0$ when $k \rightarrow \infty$

The book I am reading wants to prove that for any $\Psi \in L^2([0;T], H^1_0(\Omega))$ and any $\phi \in L^2([0;T] \times \Omega )$;

$u_k$ converges towards $u$ in the space $ \mathcal{D}'(]0;T[ \times \Omega)$, the sequence $(u_k)$ is bounded in $L^2([0;T], H^1_0(\Omega))$, so as $ \mathcal{D}(]0;T[ \times \Omega)$ is dense both in $L^2([0;T], H^1_0(\Omega))$ and in $L^2([0;T]\times \Omega)$ the weak convergence holds both in the space $L^2([0;T], H^1_0(\Omega))$ and $L^2([0;T]\times \Omega)$.

I don't understand this argument, I don't see how we could pass to the limit both for k tending to $\infty$ and for a sequence $\phi_n \in \mathcal{D}(]0;T[ \times \Omega)$ tending towards $\phi$ or $\Psi$, does any of you find some clarity in this ?

1 Answer
1

I guess the only delicate point is to retrieve (weak) convergence for the whole sequence, not only along some subsequence (which is trivial by the Banach Alaoglu theorem since $u_k$ is bounded in $L^2(0,T;H^1_0)$). A classical separation argument would do the trick here: if any converging subsequence converges to the same limit, then in fact the whole sequence converges. Here is how I think the author argues.

Fix $\Phi,\Psi\in \mathcal{D}((0,T)\times\Omega)$. Your original assumption $\int_{[0;T]\times K} |u_k(t,x)-u(t,x)|^2dtdx \rightarrow 0$ implies in particular $u_k\to u$ in $\mathcal{D}'((0,T)\times\Omega)$, so by definition of the convergence in the sense of distributions
$$
\int_{[0;T]\times \Omega} u_k(t,x) \Phi(t,x) dtdx =\left< u_k, \Phi\right>_{\mathcal{D}',\mathcal{D}}\to \left< u, \Phi\right>_{\mathcal{D}',\mathcal{D}}\hspace{2cm}(1)
$$
and
$$
\int_{[0;T]\times \Omega} \nabla u_k(t,x) \cdot \nabla \Psi(t,x) dtdx =\left<\nabla u_k,\nabla \Psi\right>_{\mathcal{D}',\mathcal{D}}\to \left<\nabla u,\nabla \Psi\right>_{\mathcal{D}',\mathcal{D}}\hspace{2cm}(2)
$$
(no need for extraction or Banach Alaoglu here, this is merely the continuity of differentiation in the sense of distributions). Since $u_k$ is bounded in $L^2(0,T;L^2)$ and $L^2(0,T;H^1_0)$ you see that both left-hand sides in (1)-(2) are controlled by $|\Phi|_{L^2(0,T;L^2)}$ and $|\nabla\Psi|_{L^2(0,T;L^2)}$ respectively, thus
$$
\left|\left< u, \Phi\right>_{\mathcal{D}',\mathcal{D}}\right|\leq C |\Phi|_{L^2(0,T;L^2)}
$$
and
$$
\left|\left<\nabla u,\nabla \Psi\right>_{\mathcal{D}',\mathcal{D}}\right|\leq C |\nabla\Psi|_{L^2(0,T;L^2)}
$$
for all $\Phi,\Psi\in \mathcal{D}((0,T)\times\Omega)$. Now by density of $\mathcal{D}((0,T)\times\Omega)$ in $L^2(0,T;L^2)$ and $L^2(0,T;H^1_0)$ one concludes that the distributions $u,\nabla u$ can be regarded as $L^2$ functions, i-e
$$
\left< u, \Phi\right>_{\mathcal{D}',\mathcal{D}}=\int_{[0;T]\times \Omega} u_k(t,x) \Phi(t,x) dtdx
$$
and
$$
\left< \nabla u, \nabla\Psi\right>_{\mathcal{D}',\mathcal{D}}=\int_{[0;T]\times \Omega} \nabla u_k(t,x) \cdot\nabla\Psi(t,x) dtdx
$$
for all $\Phi,\Psi\in \mathcal{D}((0,T)\times\Omega)$. Now by (1) and (2) you get
$$
\int_{[0;T]\times \Omega} (u_k(t,x) - u(t,x))\cdot \Phi(t,x) dtdx \rightarrow 0
$$
$$
\int_{[0;T]\times \Omega} (\nabla u_k(t,x) - \nabla u(t,x)) \cdot \nabla \Psi(t,x) dtdx \rightarrow 0
$$
for all $\Phi,\Psi\in \mathcal{D}(0,T\times\Omega)$. Using one last time the density of $\mathcal{D}((0,T)\times\Omega)$ you deduce that this holds for all $\Phi\in L^2(0,T;L^2)$ and $\Psi\in L^2(0,T;H^1_0)$ (take $\Phi_n\to\Phi$ and $\Psi_n\to \Psi$ in these spaces), QED.

The fact that $\nabla u$ belong to $L^2$ wouldn't rather be coming from the inequality $|<\nabla u , \phi>|_{\mathcal{D}',\mathcal{D}}\leq C |\phi|_{L^2(0,T;L^2(\Omega))}$ which is true because the sequence $(u_k)_k$ is bounded in $L^2(0,T; H^1_0)$.
–
incasAug 1 '14 at 12:11